Fuzzy Sets

[dony]

Hello,
I have seen a solution of the Russell's paradox, but I don't understand it. I am talking about Barber paradox. The solution was using fuzzy sets and fuzzy logic. It was as follows:
Let \(\displaystyle A_T \)be the set of men in town, who shave themselves, and \(\displaystyle A_B\) be a set of men in town who are shaved by Barber. Also let\(\displaystyle \ \ f_T\ \)be membership function of fuzzy set \(\displaystyle (A_t, f_T)\), and analogically the \(\displaystyle \ f_B\). If the X is a set of all men in town, then:\(\displaystyle \forall x \in X : f_A(x) + f_B(x) = 1\), and if \(\displaystyle x_0 \)is the Barber:
\(\displaystyle f_A(x_0) = f_B(x_0)\) and that implies that\(\displaystyle f_A(x_0) = 1/2\), which means that Barber is not fully included in neigher of sets, but he is a fuzzy member of both.

If this solution is correct, could you please explain me exacly where \(\displaystyle f_A(x) + f_B(x) = 1 \)came from. I could only agree that:\(\displaystyle max\lbrace f_A(x),f_B(x)\rbrace = 1\).
Is this solution incorrect?
Sorry if I posted it in wrong section.

 

[Ishuda]

Hello,
I have seen a solution of the Russell's paradox, but I don't understand it. I am talking about Barber paradox. The solution was using fuzzy sets and fuzzy logic. It was as follows:
Let \(\displaystyle A_T \)be the set of men in town, who shave themselves, and \(\displaystyle A_B\) be a set of men in town who are shaved by Barber. Also let\(\displaystyle \ \ f_T\ \)be membership function of fuzzy set \(\displaystyle (A_t, f_T)\), and analogically the \(\displaystyle \ f_B\). If the X is a set of all men in town, then:\(\displaystyle \forall x \in X : f_A(x) + f_B(x) = 1\), and if \(\displaystyle x_0 \)is the Barber:
\(\displaystyle f_A(x_0) = f_B(x_0)\) and that implies that\(\displaystyle f_A(x_0) = 1/2\), which means that Barber is not fully included in neigher of sets, but he is a fuzzy member of both.

If this solution is correct, could you please explain me exacly where \(\displaystyle f_A(x) + f_B(x) = 1 \)came from. I could only agree that:\(\displaystyle max\lbrace f_A(x),f_B(x)\rbrace = 1\).
Is this solution incorrect?
Sorry if I posted it in wrong section.

Well now the real answer to your question is SHE doesn't shave.

However, to be (a little) more serious, if we also say the Barber is male and, with the exception of the barber, say that \(\displaystyle f_A(x) =1\) if the man shaves himself and \(\displaystyle f_A(x) =0\) otherwise and simular for \(\displaystyle f_B(x)\), then we can say, with the exception of the barber, \(\displaystyle f_A(x)\, +\, f_B(x)\, =\, 1\). We now ask the question what is \(\displaystyle f_A(x_0)\) [and \(\displaystyle f_B(x_0)\)] where \(\displaystyle x_0\) represents the barber (and we are (semi)consistent with A & B definitions)? That is, who shaves the barber?

Actually, the answer isn't 1/2 as I have seen it. The answer is \(\displaystyle f_A(x_0)\, =\, \alpha\), \(\displaystyle 0\, \le\, \alpha\, \le\, 1\) and, of course, \(\displaystyle f_B(x_0)\, =\, 1\, -\, \alpha\), \(\displaystyle 0\, \le\, 1\, -\, \alpha\, \le\, 1\). IF the two are equal, then \(\displaystyle \alpha = \frac{1}{2}\).

 

[dony]

Actually, the answer isn't 1/2 as I have seen it. The answer is \(\displaystyle f_A(x_0)\, =\, \alpha\), \(\displaystyle 0\, \le\, \alpha\, \le\, 1\) and, of course, \(\displaystyle f_B(x_0)\, =\, 1\, -\, \alpha\), \(\displaystyle 0\, \le\, 1\, -\, \alpha\, \le\, 1\). IF the two are equal, then \(\displaystyle \alpha = \frac{1}{2}\).

Are you sure it can be 0 or 1? We got contradiction then using classical logic.

I am not sure I got it right. If as above we let be \(\displaystyle X\) a set of all men in town, and \(\displaystyle f_X \) be membership function of fuzzy set \(\displaystyle (X, f_X)\), then for sure we agree that \(\displaystyle f_X(x_0) = 1\), where \(\displaystyle x_0\) is a Barber.
Since the union of two fuzzy sets(in this example \(\displaystyle (A, f_A),(B, f_B)\)) is defined as set whose membership function is as follows:
\(\displaystyle f_\left(A\cup B\right)(x) = max \lbrace f_A(x),f_B(x) \rbrace\),
there would seem to be equality:
\(\displaystyle f_\left(A\cup B\right)(x_0) = max \lbrace f_A(x_0),f_B(x_0) \rbrace\), but I don't see it here.
It looks like it could be the result of the following:
\(\displaystyle A \cap (X\setminus A) = \emptyset\)
\(\displaystyle A \cup (X\setminus A) = X\)
are not always true (when we consider A in terms of fuzzy set).
Am I wrong?

I edited my post.

 

[Ishuda]

Are you sure it can be 0 or 1? We got contradiction then using classical logic.

I am not sure I got it right. If as above we let be \(\displaystyle X\) a set of all men in town, and \(\displaystyle f_X \) be membership function of fuzzy set \(\displaystyle (X, f_X)\), then for sure we agree that \(\displaystyle f_X(x_0) = 1\), where \(\displaystyle x_0\) is a Barber.
Since the union of two fuzzy sets(in this example \(\displaystyle (A, f_A),(B, f_B)\)) is defined as set whose membership function is as follows:
\(\displaystyle f_\left(A\cup B\right)(x) = max \lbrace f_A(x),f_B(x) \rbrace\),
there would seem to be equality:
\(\displaystyle f_\left(A\cup B\right)(x_0) = max \lbrace f_A(x_0),f_B(x_0) \rbrace\), but I don't see it here.
It looks like it could be the result of the following:
\(\displaystyle A \cap (X\setminus A) = \emptyset\)
\(\displaystyle A \cup (X\setminus A) = X\)
are not always true (when we consider A in terms of fuzzy set).
Am I wrong?

I edited my post.

The general standard rule for the union of fuzzy sets A and B is
\(\displaystyle A(x)\,\cup\, B(x)\, =\, max \lbrace A(x),B(x)\rbrace\)
with the min for intersection, see
http://www.sjsu.edu/faculty/watkins/fuzzysets.htm
for example. One also has
\(\displaystyle f_{NOT\, A}(x)\, = 1\, -\, f_A(x)\)
which is really all I meant by the "Actually, the answer isn't ..."

 

[dony]

Then what you're saying is that we have not solved this paradox. Becouse \(\displaystyle \alpha \in [0,1]\) always by definition. Shouldn't it be at least: \(\displaystyle \alpha \in (0,1)\)? If we allow 1 or 0 into the set, we get contradiction at these points, like we should. Otherwise it wouldn't be paradox in a first place. Am I wrong?

 

[Ishuda]

Then what you're saying is that we have not solved this paradox. Becouse \(\displaystyle \alpha \in [0,1]\) always by definition. Shouldn't it be at least: \(\displaystyle \alpha \in (0,1)\)? If we allow 1 or 0 into the set, we get contradiction at these points, like we should. Otherwise it wouldn't be paradox in a first place. Am I wrong?

Yes, I would accept the correction to (0,1). As far as having solved the riddle, I give you another "Is Schrodinger's cat dead or alive?"

 

[dony]

Yes, I would accept the correction to (0,1). As far as having solved the riddle, I give you another "Is Schrodinger's cat dead or alive?"

He is neither dead, nor alive. Does that mean that the cat is a fuzzy member of D(death) and L(life), when D, L are fuzzy sets?

 

[Ishuda]

He is neither dead, nor alive. Does that mean that the cat is a fuzzy member of D(death) and L(life), when D, L are fuzzy sets?

Well I would think at least as much as \(\displaystyle x_0\) is in both sets A & B. An we don't even know whether x₀ was fuzzy or not. Maybe just partially fuzzy?

 

[dony]

Well I would think at least as much as \(\displaystyle x_0\) is in both sets A & B. An we don't even know whether x₀ was fuzzy or not. Maybe just partially fuzzy?

According to definition, \(\displaystyle x_0 \) is a fuzzy member, if \(\displaystyle \alpha \in (0,1) \). What does it mean "partially fuzzy" ?

 

[Ishuda]

According to definition, \(\displaystyle x_0 \) is a fuzzy member, if \(\displaystyle \alpha \in (0,1) \). What does it mean "partially fuzzy" ?

Sorry, trying to be funny. Does he have a beard (is he fuzzy) or does he not have a beard (is not fuzzy). Like the fuzzy cat because it has hair or from the old nursery rhyme: Fuzzy Wuzzy was a bear. Fuzzy Wuzzy had no hair. So Fuzzy Wuzzy wasn't fuzzy, was he? [Fuzzy Wuzzy wasn't Fuzzy Wuzzy] So maybe x₀ only had partial hair and was partially fuzzy.

 

[dony]

Sorry, trying to be funny. Does he have a beard (is he fuzzy) or does he not have a beard (is not fuzzy). Like the fuzzy cat because it has hair or from the old nursery rhyme: Fuzzy Wuzzy was a bear. Fuzzy Wuzzy had no hair. So Fuzzy Wuzzy wasn't fuzzy, was he? [Fuzzy Wuzzy wasn't Fuzzy Wuzzy] So maybe x₀ only had partial hair and was partially fuzzy.

Well, since we got a contradiction if \(\displaystyle x_0 \) shaves himselfand the same if he is shaved by Barber, then there is noone who shaves him, so he is definitelly fuzzy.

Okay. lets be serious now. What is the meaning of the final answer?